MECHANICSKinematicsPhysics Calculator
🎯

Maximum Height

The maximum height of a projectile occurs when vertical velocity reaches zero. For launch angle θ, h_max = (v₀² sin²θ)/(2g).

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45° launch gives maximum range but not maximum height. 90° launch maximizes height for given initial speed. Time to apex equals time from apex to ground. Horizontal velocity remains constant (no air resistance).

Key quantities
h = v₀²sin²θ/(2g)
Max Height
Key relation
t = v₀sinθ/g
Time to Apex
Key relation
Max range angle
45° Optimal
Key relation
v_y = v₀sinθ − gt
Vertical v
Key relation

Ready to run the numbers?

Why: Essential for ballistics, sports physics, and understanding two-dimensional motion.

How: Kinematic equations with constant g; maximum height occurs when v_y = 0.

45° launch gives maximum range but not maximum height.90° launch maximizes height for given initial speed.

Run the calculator when you are ready.

Solve the EquationCalculate projectile maximum height and trajectory

🌍 Gravity Presets

🔧 Calculation Mode

⚙️ Input Parameters

📊 Results

Maximum Height
10.19
meters
Time to Apex
1.442
seconds
Total Flight Time
2.88
seconds
Horizontal Range
40.77
meters
Height Gain
10.19 m
v₀ᵧ (vertical)
14.14 m/s
v₀ₓ (horizontal)
14.14 m/s
Impact Velocity
20.00 m/s

📈 Visualizations

Projectile Trajectory

Height vs Time

Height Breakdown

📝 Step-by-Step Solution

📊 Input Parameters

Gravity: g = 9.81 m/s²

Initial height: h₀ = 0.00 m

Initial velocity: v₀ = 20.00 m/s

Launch angle: θ = 45.0°

🎯 Velocity Components

Vertical component: v₀ᵧ = v₀ × sin(θ)

v₀ᵧ = 20.0000 × sin(45.0000°)

→ v₀ᵧ = 14.14 m/s

Horizontal component: v₀ₓ = v₀ × cos(θ)

v₀ₓ = 20.0000 × cos(45.0000°)

→ v₀ₓ = 14.14 m/s

📏 Maximum Height Calculation

Time to reach apex: t_apex = v₀ᵧ / g

t_apex = 14.1421 / 9.8100

→ t_apex = 1.442 seconds

Height gain: Δh = v₀ᵧ² / (2g)

Δh = 14.1421² / (2 × 9.8100)

→ Δh = 10.19 m

Maximum Height: H_max = h₀ + Δh

H_max = 0.0000 + 10.1937

→ H_max = 10.19 m

🛬 Complete Flight Analysis

Horizontal distance at apex

→ x_apex = 20.39 m

Total flight time

→ t_total = 2.883 s

Total horizontal range

→ R = 40.77 m

Impact velocity

→ v_impact = 20.00 m/s

📖 Maximum Height Formula

The maximum height of a projectile is the highest point it reaches during flight. At this point, the vertical velocity is zero.

H_max = h₀ + v₀ᵧ² / (2g) = h₀ + (v₀ sin θ)² / (2g)

Maximum height = Initial height + Height gained

At the Apex

  • • Vertical velocity = 0
  • • Horizontal velocity unchanged
  • • Maximum potential energy
  • • Minimum kinetic energy

Time to Apex

t_apex = v₀ᵧ / g = v₀ sin(θ) / g

Half the total flight time (for launch from ground)

Key Insight

Maximum height depends only on vertical velocity component and gravity. Horizontal velocity has no effect on height reached.

📐 Effect of Launch Angle

Anglesin(θ)Height FactorRange FactorBest For
15°0.267%50%Low drives
30°0.5025%87%Long throws
45°0.7150%100%Max range
60°0.8775%87%High arcs
90°1.00100%0%Max height

Height factor = sin²(θ), Range factor = sin(2θ) (relative to maximum)

Frequently Asked Questions

Q: Why is 90° the angle for maximum height?

At 90°, all the initial velocity goes into the vertical component (sin 90° = 1). Any other angle splits velocity between horizontal and vertical, reducing the vertical component.

Q: Does horizontal velocity affect maximum height?

No! Horizontal and vertical motions are independent. A ball thrown straight up at 10 m/s reaches the same height as one thrown at 10 m/s vertically with any horizontal velocity.

Q: How does air resistance affect maximum height?

Air resistance always reduces maximum height compared to the ideal case. The projectile loses energy to drag on the way up, reaching a lower apex. The effect is larger for light, fast objects.

📚 Key Takeaways

Essential Formulas

  • ✓ H_max = h₀ + v₀ᵧ²/(2g)
  • ✓ t_apex = v₀ᵧ/g
  • ✓ At apex: vᵧ = 0
  • ✓ v₀ᵧ = v₀ × sin(θ)
  • ✓ 90° gives maximum height

Practical Tips

  • ✓ Higher angle = more height, less range
  • ✓ 45° balances height and range
  • ✓ Initial height adds directly to max height
  • ✓ Energy converts between KE and PE
  • ✓ Air resistance reduces all values

Energy Analysis

Maximum height can also be derived using conservation of energy:

Energy Method

At launch: KE = ½mv₀ᵧ², PE = mgh₀

At apex: KE = 0, PE = mgH_max

Conservation: KE₀ + PE₀ = PE_max

½mv₀ᵧ² + mgh₀ = mgH_max

H_max = h₀ + v₀ᵧ²/(2g)

Energy Distribution

  • At launch: Max KE (vertical), some PE if elevated
  • Rising: KE converting to PE, vᵧ decreasing
  • At apex: Zero vertical KE, max PE achieved
  • Falling: PE converting back to KE
  • Note: Horizontal KE stays constant throughout

🌍 Maximum Heights on Different Planets

Same initial conditions (v₀ = 20 m/s, θ = 90°) produce different maximum heights:

Planet/Moong (m/s²)Max HeightTime to Apexvs Earth
🌍 Earth9.8120.4 m2.04 s1.0×
🌙 Moon1.62123.5 m12.35 s6.1×
🔴 Mars3.7153.9 m5.39 s2.6×
🪐 Venus8.8722.5 m2.25 s1.1×
🟠 Jupiter24.798.1 m0.81 s0.4×
🧊 Europa1.31152.7 m15.27 s7.5×

Lower gravity = higher maximum height for the same initial velocity

🏆 World Records & Extremes

Sports Records

  • High Jump: 2.45 m (Javier Sotomayor) - requires ~7.0 m/s vertical
  • Pole Vault: 6.21 m (Armand Duplantis) - uses pole as energy transfer
  • Long Jump: 8.95 m (Mike Powell) - max height ~1.3 m at 45° equivalent
  • Shot Put: 23.37 m - release height ~2.1 m, apex ~4.5 m

Engineering Feats

  • Artillery shell: Max altitude ~25 km (Paris Gun, WWI)
  • Rocket: SpaceX Starship reaches ~100+ km
  • Baseball (theoretical): ~100 m if thrown at 45 m/s straight up
  • Golf ball: ~50 m apex on professional drives

🧪 Experimental Verification

Methods to verify maximum height calculations in physics labs:

Video Analysis

  1. Record projectile with high-speed camera
  2. Include meter stick for scale reference
  3. Import video to Tracker or LoggerPro
  4. Mark position frame-by-frame
  5. Find frame where vertical motion stops

Time-Based Method

  1. Measure total flight time (t_total)
  2. For symmetric flight: t_apex = t_total/2
  3. Calculate: H = ½g(t_apex)²
  4. Add initial launch height
  5. Compare to direct measurement

🎯 Real-World Applications

Sports

  • • Basketball shot arc optimization
  • • Football punt hang time
  • • Baseball pop fly height
  • • Volleyball serve trajectory
  • • Diving platform clearance

Engineering

  • • Fountain water jet design
  • • Fireworks display timing
  • • Irrigation sprinkler range
  • • Amusement ride safety
  • • Ballistic missile defense

Everyday Life

  • • How high can I throw a ball?
  • • Water balloon trajectories
  • • Ceiling height for juggling
  • • Trampoline jump heights
  • • Kite flying altitude

⚠️ Common Mistakes to Avoid

❌ Calculation Errors

  • • Using total velocity instead of vertical component
  • • Forgetting to add initial height h₀
  • • Using degrees instead of radians for sin(θ)
  • • Wrong sign for gravity (should be positive in formula)
  • • Confusing maximum height with horizontal range

✓ Best Practices

  • • Always extract vertical component: v₀ᵧ = v₀sin(θ)
  • • Include launch height in final answer
  • • Check calculator angle mode (DEG vs RAD)
  • • Remember at apex: vᵧ = 0 (not v = 0!)
  • • Verify answer makes physical sense

💡 Problem-Solving Strategy

  1. 1Identify what's given: Initial velocity, angle, and/or launch height
  2. 2Find vertical velocity: v₀ᵧ = v₀ × sin(θ). For vertical throws, v₀ᵧ = v₀
  3. 3Apply the formula: H_max = h₀ + v₀ᵧ²/(2g)
  4. 4Calculate time to apex: t_apex = v₀ᵧ/g (if needed)
  5. 5Verify: Check that height is reasonable for the given speed

📐 Mathematical Derivation

From Kinematics (v² = v₀² + 2aΔy)

At maximum height, vᵧ = 0 (momentarily stationary)

Using: vᵧ² = v₀ᵧ² - 2g(H_max - h₀)

0 = v₀ᵧ² - 2g(H_max - h₀)

2g(H_max - h₀) = v₀ᵧ²

H_max - h₀ = v₀ᵧ²/(2g)

H_max = h₀ + v₀ᵧ²/(2g) ✓

🎮 Fun Physics Scenarios

Explore these interesting maximum height scenarios:

🌙 Apollo Astronaut Golf

Alan Shepard hit golf balls on the Moon (Apollo 14). With g = 1.62 m/s² and no air resistance:

At 45° and 40 m/s initial velocity:

H_max = (40×sin45)²/(2×1.62) = 246 m

(vs 40.8 m on Earth)

🌋 Volcanic Eruption

Volcanic bombs can be ejected at incredible speeds:

At 400 m/s vertical:

H_max = 400²/(2×9.81) = 8.15 km!

(Air resistance reduces this significantly)

🏀 Perfect Free Throw

A basketball free throw with optimal arc of 45°:

Release at 2m, target at 3.05m, v = 7 m/s:

v₀ᵧ = 7×sin(52°) = 5.5 m/s

H_max = 2 + 5.5²/19.6 = 3.5 m

💧 Water Fountain

Designing a fountain with 5m high jets:

Required velocity: v = √(2gh)

v = √(2×9.81×5) = 9.9 m/s

At the nozzle exit

📊 Height vs Velocity Comparison

Since H ∝ v², doubling velocity quadruples maximum height:

Vertical VelocityMax Height (Earth)Time to ApexExample
5 m/s1.27 m0.51 sLight toss
10 m/s5.10 m1.02 sThrow overhead
15 m/s11.47 m1.53 sStrong throw
20 m/s20.39 m2.04 sBaseball pitch (up)
30 m/s45.87 m3.06 sCatapult
50 m/s127.4 m5.10 sCannon

Note: Doubling v₀ᵧ quadruples H_max (since H ∝ v²)

🎓 Practice Problems

Problem 1: Basketball Jump Shot

A player releases a basketball at 2.5m height with initial velocity 8 m/s at 55° above horizontal. Find the maximum height of the ball.

v₀ᵧ = 8 × sin(55°) = 6.55 m/s

H_max = 2.5 + (6.55)²/(2×9.81)

H_max = 2.5 + 2.19 = 4.69 m

Problem 2: Straight Up Throw

A ball is thrown straight up with initial velocity 25 m/s from ground level. How high does it go? How long until it reaches the apex?

H_max = 0 + (25)²/(2×9.81) = 31.9 m

t_apex = 25/9.81 = 2.55 s

Problem 3: Required Velocity

What vertical velocity is needed to reach a height of 50 meters?

50 = v₀ᵧ²/(2×9.81)

v₀ᵧ² = 50 × 19.62 = 981

v₀ᵧ = √981 = 31.3 m/s

🌬️ Air Resistance Effects

In reality, air resistance reduces maximum height, especially for fast or light objects:

What Air Resistance Does

  • • Always opposes motion (slows projectile on way up)
  • • Reduces maximum height achieved
  • • Time to apex is shorter than calculated
  • • Energy lost to drag work
  • • Effect increases with velocity (F_drag ∝ v²)

When to Consider It

  • • High velocities (>30 m/s)
  • • Light objects (feathers, badminton)
  • • Large surface area relative to mass
  • • Precision calculations needed
  • • Real ballistics applications

For most educational problems, ignoring air resistance gives results within 5-10% of actual values at typical sports velocities.

Frequently Asked Questions

Q: Does mass affect maximum height?

Without air resistance, no! H_max = v₀²sin²(θ)/(2g) has no mass term. All objects launched at the same speed and angle reach the same height (in vacuum). With air resistance, lighter objects are affected more.

Q: Why is maximum height reached at the midpoint of flight?

The trajectory is symmetric (parabolic). Vertical velocity decreases at constant rate g, reaching zero at peak, then increases in the downward direction at the same rate. Time up = time down.

📝 Key Takeaways

  • • Maximum height formula: H_max = v₀²sin²(θ)/(2g)
  • • For vertical launch (θ = 90°): H_max = v₀²/(2g)
  • • At peak, vertical velocity = 0 (horizontal unchanged)
  • • Time to peak: t = v₀sin(θ)/g
  • • Mass does not affect height (without air resistance)
  • • 90° gives maximum possible height for any speed

🔢 Speed vs Height Quick Reference

10 m/s → 5.1 m max

20 m/s → 20.4 m max

30 m/s → 45.9 m max

H ∝ v² (double v = 4× height)

45° angle: H = ¼ × Range

💡 Quick Tip

For maximum range (horizontal distance), use 45°. For maximum height, use 90° (straight up). At 45°, the maximum height is exactly 1/4 of the range.

For educational and informational purposes only. Verify with a qualified professional.

🔬 Physics Facts

🎯

Maximum height h = v₀²sin²θ/(2g) occurs at t = v₀sinθ/g.

— Halliday

📐

45° launch angle gives maximum range for level ground.

— HyperPhysics

⏱️

Time to apex equals time from apex to ground (symmetry).

— Serway

📊

At apex, vertical velocity is zero; horizontal velocity unchanged.

— NIST

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